On Fri, 7 Nov 2003, Cameron Purchase wrote: ...>... I can calculate> the sine of any angle by the well known series:>> X - X^3/3! + X^5/5!......>> but how would I go about doing this in reverse so to speak? That is> to say if I had the ratios of the right angled triangle and no> recourse to a calculator with trigometrical functions or tables how> would I work out the angle? How would I convert the .5 side of the> triangle to the .523 of its radian measure?> ...

You could use the less well-known, but analogous, series for the inversesine

arcsin(x) = x + s^3/6 + 3x^5/40 + ...

just using the first three terms we get

arcsin(.5) = .523177...

which is pretty close to the actual value .523598...

The series above for arcsin came from the same place that you got theseries for sine -- the Taylor series expansion, which you can find in anycalculus book.